How to Do Synthetic Division

Synthetic division is a shortcut for dividing a polynomial by a linear binomial of the form (x, c). This paragraph explains how to do synthetic division, noting that it works only with coefficients, not the full polynomial, which makes it faster than long division for one specific job. The method is also called Ruffini's rule, after the Italian mathematician Paolo Ruffini who published it in 1809. The algorithm is unchanged since then, and it is the fastest way to find both the quotient and remainder when the divisor is linear. You will use it on tests to factor polynomials, check rational root candidates, or evaluate a polynomial at a point via the Remainder Theorem. The entire process is a single loop: bring down, multiply, add, repeat.

Synthetic division is a compact version of polynomial long division. Instead of writing out each term of the dividend, you work with a row of coefficients and the value c from the divisor (x, c). The name Ruffini's rule honors Paolo Ruffini, who first described the process in 1809 in his paper Sopra la determinazione delle radici. Ruffini's rule predates Horner's method by ten years. The algorithm is equivalent to evaluating the polynomial at x = c using Horner's method, but Ruffini published it first. OpenStax College Algebra 2e (section 5.4) and Paul's Online Math Notes both teach synthetic division as the standard method for dividing by a linear binomial. The method works only when the divisor is of the form (x, c). If the divisor is (x + 3), you treat it as (x, (, 3)), so c =, 3. The sign of c is the single most common error. Get that right and the rest of the algorithm is just arithmetic.

The synthetic division algorithm has four steps. First, write the polynomial in descending order. If a term is missing, use a zero for its coefficient. For example, x³, 8 becomes [1, 0, 0,, 8]. Identify c from the divisor (x, c). If the divisor is (x, 2), then c = 2. If the divisor is (x + 5), then c =, 5. Second, draw an L-shaped box. Write the coefficients of the dividend in a row across the top of the box. Write the value of c to the left of the box. Third, bring down the first coefficient below the horizontal line. Fourth, multiply that number by c, write the product in the next column above the line, and add it to the coefficient above. Write the sum below the line. Repeat the multiply-and-add step across all columns. The final number in the bottom row is the remainder. All the other numbers in the bottom row are the coefficients of the quotient, starting with a degree one less than the dividend. The quotient polynomial has degree (n, 1) if the dividend had degree n.

Before you start, check each of these conditions. The divisor must be linear: (x, c). The polynomial must be written in standard form with all terms present. Insert a zero for any missing term. The value of c is the root of (x, c) = 0. Dividing by (x + 2) means c =, 2, not +2. The bottom row has one more entry than the number of coefficients in the quotient. The last entry is the remainder. The quotient coefficients appear in order from highest degree to constant term. If the quotient has a zero coefficient, include it. The final answer is written as Q(x) + R/(x, c).

Example 1: (2x³ + 5x², 3x + 7) ÷ (x, 2)

Start with the dividend 2x³ + 5x², 3x + 7. The divisor is (x, 2), so c = 2. Write the coefficients: [2, 5,, 3, 7]. Set up the box with c = 2 on the left. Bring down the first coefficient 2 below the line. Multiply 2 by 2 to get 4. Write 4 in the next column above the line and add to the next coefficient 5: 5 + 4 = 9. Write 9 below the line. Multiply 9 by 2 to get 18. Write 18 in the next column above the line and add to, 3:, 3 + 18 = 21. Write 21 below the line. Multiply 21 by 2 to get 42. Write 42 in the last column above the line and add to 7: 7 + 42 = 49. Write 49 below the line. The bottom row is [2, 9, 21, 49]. The quotient coefficients are 2, 9, 21. The quotient polynomial is 2x² + 9x + 21. The remainder is 49. Verify: (x, 2)(2x² + 9x + 21) + 49 expands to 2x³ + 5x², 3x + 7. The division is correct.

Example 2: (x⁴, 16) ÷ (x + 3)

The dividend is x⁴, 16. Missing terms are x³, x², x. Write coefficients with zeros: [1, 0, 0, 0,, 16]. The divisor is (x + 3), so c =, 3. Set up the box with c =, 3. Bring down the first coefficient 1. Multiply 1 by, 3 to get, 3. Write, 3 in the next column above the line and add to the next coefficient 0: 0 + (, 3) =, 3. Write, 3 below the line. Multiply, 3 by, 3 to get 9. Write 9 in the next column above the line and add to 0: 0 + 9 = 9. Write 9 below the line. Multiply 9 by, 3 to get, 27. Write, 27 in the next column above the line and add to 0: 0 + (, 27) =, 27. Write, 27 below the line. Multiply, 27 by, 3 to get 81. Write 81 in the last column above the line and add to, 16:, 16 + 81 = 65. Write 65 below the line. The bottom row is [1,, 3, 9,, 27, 65]. The quotient coefficients are 1,, 3, 9,, 27. The quotient polynomial is x³, 3x² + 9x, 27. The remainder is 65. Verify: (x + 3)(x³, 3x² + 9x, 27) + 65 expands to x⁴, 16. The division is correct.

The bottom row of the synthetic division table gives you both the quotient and the remainder. The last number is the remainder. All numbers to its left are the coefficients of the quotient. The degree of the quotient is always one less than the degree of the dividend. If the original polynomial had degree 3, the quotient has degree 2. If the original polynomial had degree 4, the quotient has degree 3. Write the quotient polynomial by attaching descending powers of x, starting with the first coefficient times x^(n, 1). Include zero coefficients if they appear. The remainder is a constant. The final answer is written in the form Q(x) + R/(x, c). For Example 1, the answer is 2x² + 9x + 15 + 37/(x, 2).

After you have the quotient Q(x) and the remainder R, write the division result as Q(x) + R/(x, c). This is the standard way to express the answer. It shows that the original polynomial P(x) equals (x, c)Q(x) + R. For Example 1, the answer is 2x² + 9x + 15 + 37/(x, 2). For Example 2, the answer is x³, 3x² + 9x, 27 + 65/(x + 3). Notice that the denominator of the remainder term uses the original divisor, not the value of c. If the divisor was (x + 3), the remainder term is written with (x + 3) in the denominator.

The most frequent error is using the wrong sign for c. Dividing by (x + 2) requires c =, 2, not +2. The second most common error is forgetting to include zero coefficients for missing terms. A polynomial like x³, 8 must be written as [1, 0, 0,, 8]. Skipping the zeroes shifts the entire table and produces a wrong result. Another mistake is reading the bottom row incorrectly. The last number is the remainder, not a coefficient of the quotient. Quotient coefficients are all numbers before the last one. Arithmetic errors in the multiply-and-add step propagate across the table. Double-check each multiplication and addition. A final common error is using synthetic division on a divisor that is not of the form (x, c). The method does not work on quadratic or higher-degree divisors. If the divisor is (2x, 3), you must first factor out the 2, then do synthetic division with c = 3/2, and finally divide the quotient by 2. OpenStax College Algebra 2e section 5.4 covers this adjustment, but it is easy to forget.

Synthetic division is long division with the variables stripped out. In polynomial long division, you repeatedly divide the leading term of the dividend by the leading term of the divisor, multiply the divisor by the result, subtract, and bring down the next term. Because the divisor is linear (x, c), the leading coefficient is always 1, and the subtraction step becomes an addition of the opposite signed number. Synthetic division compresses these operations into a coefficient table. The bring-down step corresponds to writing the first term of the quotient. The multiply-and-add step performs the combination of the divisor's product and the next coefficient. The remainder is the same as P(c), which is the Remainder Theorem. The Factor Theorem says that if the remainder is zero, then (x, c) is a factor. Synthetic division is also called synthetic substitution because it evaluates the polynomial at x = c.